Find Entire Functions for f(2-i)=4i and |f(z)|≤e^2

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Homework Help Overview

The discussion revolves around finding entire functions that satisfy the conditions f(2-i) = 4i and |f(z)| ≤ e^2. The problem is situated within the context of complex analysis, particularly focusing on properties of entire functions and the implications of boundedness.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply a theorem regarding the growth of entire functions and questions how to determine the degree of the polynomial. Some participants inquire whether the boundedness condition applies to all z and reference Liouville's theorem in relation to the problem.

Discussion Status

The discussion is active, with participants exploring the implications of the theorems mentioned. There is recognition of the relationship between the original poster's theorem and Liouville's theorem, with some participants suggesting that the original poster's understanding may need clarification. No consensus has been reached regarding the best approach to the problem.

Contextual Notes

Participants note the importance of understanding the conditions under which the theorems apply, particularly in relation to the boundedness of entire functions. There is an ongoing exploration of the definitions and implications of the theorems discussed.

jaci55555
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find all the entire functions f for which f(2-i) = 4i and |f(z)|<= e^2


if f is entire on Complex, and there exists M,K element R^+ and k element N such that
|f(z)|<= M * |z^k|
for all z element Complex with |z|>=K then f is polynomial with degree <=k


can i use this thereom? if i do, then the |z^k| part is 1 and so M is e^2.
but then i should try find the k...
I'm not sure

i know that x^2 - 6X + 9 is an entire function there, but i just worked it out...
 
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jaci55555 said:
find all the entire functions f for which f(2-i) = 4i and |f(z)|<= e^2

Do you mean that this holds for all z?
Do you know Liouville's theorem?
 
The OP gave a stronger theorem than Liouville, which is the case k = 0 in the above theorem.
 
snipez90 said:
The OP gave a stronger theorem than Liouville, which is the case k = 0 in the above theorem.

Yes, I'm aware. But Liouville is enough here. Stronger theorems somethimes obfusciate things...
 
micromass said:
Do you mean that this holds for all z?
Do you know Liouville's theorem?

Thanks for answering :)
That is all the question stated - so i think it does hold for all z.
I thought the theorem i stated above WAS Liouville's theorem?
But i wasn't sure about how to use it.
 
Yes, of course the theorem you stated is Liouville's theorem. But the theorem originally known as Liouville's is "every entire, bounded function is constant". This follows easily from the theorem you stated (try to prove this!)
 
thank you
 

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